brightlightkim

Given two binary strings a and b, return their sum as a binary string. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010", b = "1011" Output: "10101" Constraints: 1 =0){ if(i>=0){ sum = sum + a.charAt(i) - '0'; i--; } if(j>=0){ sum = sum + b.charAt(j) - '0'; j--; } sb.append(sum%2); sum = sum/2; } if(sum!=0) sb.append(sum); return sb.reverse().toString(); } }

Bit Manipulation by Hand: ^ means XOR, ~ is a Not (Negation) - XOR: If it's a different bit, it will be 1. EX) 1101 ^ (~1101) = 1111 - If you XOR a bit with its own negated value, you will always get 1. Therefore, the solution to a^(~a) will be a sequence of 1s. EX) 1011 & (~0